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Mathematical Excalibur, Vol. 8, No. 4, Aug 03- Oct 03
RECEN 和 IMCEN 用來定出中心 點的位置，現在以 (0, 0) 為中心（見 第 30 行）。我們可以通過更改 R、 RECEN 和 IMCEN 的值來移動或放 大曼德勃羅集。 第 40 行選擇繪圖的模式及清除 舊有的畫面。 程序的第 50 及 60 行定出畫面上 的坐標 X 和 Y，然後在第 70 行計 算出對應複數 c0 的實值和虛值。 注意：若 c0 = a0 + b0 i，cn = an + bn i， 則 cn+1 = cn2 + c0 = (an + bn i)2 + (a0 + b0 i) = an2 − bn2 + 2anbn i + a0 + b0 i = (an2 − bn2 + a0) + (2anbn + b0)i。 所以 cn+1 的實部等於 an2 − bn2 + a0， 而虛部則等於 2anbn + b0。 將以上的計算化成程序，得第 110 及 120 行。REC 和 IMC 分別是 c0 的實值和虛值。RE 和 IM 分別是 cn 的實值和虛值。RE2 和 IM2 分別 是 cn 的實值和虛值的平方。
IMO 2003 T. W. Leung The 44th International Mathematical Olympiad (IMO) was held in Tokyo, Japan during the period 7 - 19 July 2003. Because Hong Kong was declared cleared from SARS on June 23, our team was able to leave for Japan as scheduled. The Hong Kong Team was composed as follows. Chung Tat Chi (Queen Elizabeth School) Kwok Tsz Chiu (Yuen Long Merchants Assn. Sec. School) Lau Wai Shun (T. W. Public Ho Chuen Yiu Memorial College) Siu Tsz Hang (STFA Leung Kau Kui College) Yeung Kai Sing (La Salle College) Yu Hok Pun (SKH Bishop Baker Secondary School) Leung Tat Wing (Leader) Leung Chit Wan (Deputy Leader) Two former Hong Kong Team members, Poon Wai Hoi and Law Ka Ho, paid us a visit in Japan during this period.
執行本程序所須的時間，要視乎 電腦的速度，以現時一般的電腦而 言，整個程序應該可以 1 分鐘左右 完成。
The contestants took two 4.5 Hours contests on the mornings of July 13 and 14. Each contest consisted of three questions, hence contest 1 composed of Problem 1 to 3, contest 2 Problem 4 to 6. In each contest usually the easier problems come first and harder ones come later. After normal coordination procedures and Jury meetings cutoff scores for gold, silver and bronze medals were decided. This year the cutoff scores for gold, silver and bronze medals were 29, 19 and 13 respectively. Our team managed to win two silvers, two bronzes and one honorable mention. (Silver: Kwok Tsz Chiu and Yu Hok Pun, Bronze: Siu Tsz Hang and Yeung Kai Sing, Honorable Mention: Chung Tat Chi, he got a full score of 7 on one question, which accounted for his honorable mention, and his total score is 1 point short of bronze). Among all contestants three managed to obtain a perfect score of 42 on all six questions. One contestant was from China and the other two from Vietnam.
參考書目 Heinz-Otto Peitgen, Hartmut Jürgens and Dietmar Saupe (1992) Fractals for the Classroom Part Two: Introduction to Fractals and Chaos. NCTM, Springer-Verlag.
The Organizing Committee did not give official total scores for individual countries, but it is a tradition that scores between countries were compared. This year the top five teams were Bulgaria, China, U.S.A., Vietnam and Russia
J 用來紀錄第 100 至 140 行的循 環的次數。第 100 行亦同時計算 cn 模的平方。若模的平方大於 256 或 者循環次數多於 15，循環將會終 止。這時候，J 的數值越大，表示 該數列較「收斂」，即經過多次計 算後，cn 的模仍不會變得很大。第 150 至 200 行以顏色將收斂情況分 類，紅色表示最「收歛」的複數， 其次是白色，跟著是綠色、藍色和 黃色，而最快擴散的部分以黑色表 示。第 210 行以先前選定的顏色畫 出該點。 曼德勃羅集繪畫完成後，以白色 畫出橫軸及縱軸（見第 240 至 260 行） ，以供參考。程序亦在此結束。
respectively. The Bulgarian contestants did extremely well on the two hard questions, Problem 3 and 6. Many people found it surprising. On the other hand, despite going through war in 1960s Vietnam has been strong all along. Perhaps they have participated in IMOs for a long time and have a very good Russian tradition. Among 82 teams, we ranked unofficially 26. We were ahead of Greece, Spain, New Zealand and Singapore, for instance. Both New Zealand and we got our first gold last year. But this year the performance of the New Zealand Team was a bit disappointing. On the other hand, we were behind Canada, Australia, Thailand and U.K.. Australia has been doing well in the last few years, but this year the team was just 1 point ahead of us. Thailand has been able to do quite well in these few years. IMO 2004 will be held in Greece, IMO 2005 in Mexico, IMO 2006 in Slovenia. IMO 2007 will be held in Vietnam, the site was decided during this IMO in Japan. For the reader who will try out the IMO problems this year, here are some comments on Problem 3, the hardest problem in the first day of the competitions. Problem 3. A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is 3 / 2 times the sum of their lengths. Prove that all the angles of the hexagon are equal. (A convex hexagon ABCDEF has three pairs of opposite sides: AB and DE, BC and EF, CD and FA.) The problem is hard mainly because one does not know how to connect the given condition with that of the interior angles. Perhaps hexagons are not as rigid as triangles. It also reminded me of No. 5, IMO 1996, another hard problem of polygons. The main idea is as follows. Given a hexagon ABCDEF, connect AD, BE and CF to form the diagonals. From the given condition of the hexagon, it can be proved that the triangles formed by the diagonals and the sides are actually equilateral triangles. Hence the interior angles of the hexagons are 120 o . Good luck.